R 评价二分二元参数一致性的一种整齐方法

pacman::p_load(tidyverse, glue)

name_vals <- c("negative", "positive")

# join the data to itself, so every id, method, result row can see every other method and result with the same id
full_join(my_df, my_df, by = "ID", relationship = "many-to-many") |>
  count(METHOD.x, METHOD.y, RESULT.x, RESULT.y) |>

  # we need to complete the data, because the empty values in our final table have to be created using glue, using values_fiill doesn't work
  complete(METHOD.x, METHOD.y, RESULT.x, RESULT.y, fill = list(n = 0)) |>

  # create all the columns we will need for pivoting
  mutate(condition = glue("if {METHOD.x} is {name_vals[RESULT.x + 1]}"),
        name = glue("{METHOD.y} {name_vals[RESULT.y + 1]}"),
        value = glue("{round(n * 100 / max(sum(n), 1), 2)}% ({n}/{sum(n)})"), # max(sum(n), 1) because when a condition doesn't exist in our dataset, (for example, M3 positive in someone who also had M2), sum(n) == 0, dividing by zero gives us NaNs, which you probably don't want
        .by = c(METHOD.x, METHOD.y, RESULT.x))|>
  pivot_wider(id_cols = condition)

输出:

# A tibble: 6 × 7
  condition         `M1 negative` `M1 positive` `M2 negative` `M2 positive` `M3 negative` `M3 positive`
  <glue>            <glue>        <glue>        <glue>        <glue>        <glue>        <glue>       
1 if M1 is negative 100% (1/1)    0% (0/1)      100% (1/1)    0% (0/1)      100% (1/1)    0% (0/1)     
2 if M1 is positive 0% (0/2)      100% (2/2)    0% (0/1)      100% (1/1)    50% (1/2)     50% (1/2)    
3 if M2 is negative 100% (1/1)    0% (0/1)      100% (1/1)    0% (0/1)      100% (1/1)    0% (0/1)     
4 if M2 is positive 0% (0/1)      100% (1/1)    0% (0/1)      100% (1/1)    100% (1/1)    0% (0/1)     
5 if M3 is negative 50% (1/2)     50% (1/2)     50% (1/2)     50% (1/2)     100% (2/2)    0% (0/2)     
6 if M3 is positive 0% (0/1)      100% (1/1)    0% (0/0)      0% (0/0)      0% (0/1)      100% (1/1)